3.1024 \(\int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=255 \[ -\frac {35 i a^{9/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

-35*I*a^(9/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/c^(3/2)/f+35/2*I*a^4*(
a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c^2/f+35/6*I*a^3*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^
(3/2)/c^2/f+14/3*I*a^2*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(1/2)-2/3*I*a*(a+I*a*tan(f*x+e))^(7/2)/
f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.21, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ -\frac {35 i a^{9/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(9/2)/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

((-35*I)*a^(9/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(3/2)*f
) - (((2*I)/3)*a*(a + I*a*Tan[e + f*x])^(7/2))/(f*(c - I*c*Tan[e + f*x])^(3/2)) + (((14*I)/3)*a^2*(a + I*a*Tan
[e + f*x])^(5/2))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (((35*I)/2)*a^4*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*T
an[e + f*x]])/(c^2*f) + (((35*I)/6)*a^3*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(c^2*f)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{7/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {\left (35 a^3\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{3 c f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {\left (35 a^4\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {\left (35 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}-\frac {\left (35 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{c f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}-\frac {\left (35 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{c f}\\ &=-\frac {35 i a^{9/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 13.27, size = 386, normalized size = 1.51 \[ \frac {\cos ^4(e+f x) (a+i a \tan (e+f x))^{9/2} \left (\cos (2 f x) \left (\frac {32 \sin (2 e)}{3 c^2}+\frac {32 i \cos (2 e)}{3 c^2}\right )+\sin (2 f x) \left (-\frac {32 \cos (2 e)}{3 c^2}+\frac {32 i \sin (2 e)}{3 c^2}\right )-\sec (e) \sin (f x) \left (\frac {\cos (4 e)}{2 c^2}-\frac {i \sin (4 e)}{2 c^2}\right ) \sec (e+f x)+\sec (e) (36 \cos (e)+i \sin (e)) \left (\frac {\sin (4 e)}{2 c^2}+\frac {i \cos (4 e)}{2 c^2}\right )+\frac {4 \sin (4 f x)}{3 c^2}-\frac {4 i \cos (4 f x)}{3 c^2}\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^4}-\frac {35 i \sqrt {e^{i f x}} e^{-i (5 e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{9/2}}{c f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sec ^{\frac {9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(9/2)/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

((-35*I)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))]*(a + I*a*Tan[
e + f*x])^(9/2))/(c*E^(I*(5*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(9/2)*(Cos[f*x] + I*Sin
[f*x])^(9/2)) + (Cos[e + f*x]^4*((((-4*I)/3)*Cos[4*f*x])/c^2 + Cos[2*f*x]*((((32*I)/3)*Cos[2*e])/c^2 + (32*Sin
[2*e])/(3*c^2)) + Sec[e]*(36*Cos[e] + I*Sin[e])*(((I/2)*Cos[4*e])/c^2 + Sin[4*e]/(2*c^2)) - Sec[e]*Sec[e + f*x
]*(Cos[4*e]/(2*c^2) - ((I/2)*Sin[4*e])/c^2)*Sin[f*x] + ((-32*Cos[2*e])/(3*c^2) + (((32*I)/3)*Sin[2*e])/c^2)*Si
n[2*f*x] + (4*Sin[4*f*x])/(3*c^2))*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x
])^(9/2))/(f*(Cos[f*x] + I*Sin[f*x])^4)

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fricas [B]  time = 0.50, size = 426, normalized size = 1.67 \[ -\frac {105 \, \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac {8 \, {\left (a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (4 i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, c^{2} f\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 105 \, \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac {8 \, {\left (a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (-4 i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{2} f\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 2 \, {\left (-16 i \, a^{4} e^{\left (7 i \, f x + 7 i \, e\right )} + 112 i \, a^{4} e^{\left (5 i \, f x + 5 i \, e\right )} + 350 i \, a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + 210 i \, a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(105*sqrt(a^9/(c^3*f^2))*(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)*log((8*(a^4*e^(3*I*f*x + 3*I*e) + a^4*e^(I*
f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(a^9/(c^3*f^2))*(4*I*c^2
*f*e^(2*I*f*x + 2*I*e) - 4*I*c^2*f))/(a^4*e^(2*I*f*x + 2*I*e) + a^4)) - 105*sqrt(a^9/(c^3*f^2))*(c^2*f*e^(2*I*
f*x + 2*I*e) + c^2*f)*log((8*(a^4*e^(3*I*f*x + 3*I*e) + a^4*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))
*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(a^9/(c^3*f^2))*(-4*I*c^2*f*e^(2*I*f*x + 2*I*e) + 4*I*c^2*f))/(a^4*e^
(2*I*f*x + 2*I*e) + a^4)) - 2*(-16*I*a^4*e^(7*I*f*x + 7*I*e) + 112*I*a^4*e^(5*I*f*x + 5*I*e) + 350*I*a^4*e^(3*
I*f*x + 3*I*e) + 210*I*a^4*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)
))/(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {9}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(9/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)

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maple [A]  time = 0.27, size = 409, normalized size = 1.60 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{4} \left (315 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +27 i \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+105 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c -3 \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )-105 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -393 i \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -259 \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )+164 \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{6 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{3} \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^4/c^2*(315*I*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*
x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c+27*I*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*
x+e)^3+105*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^3*a*c-3*(c*a)^
(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^4-105*I*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1
/2))/(c*a)^(1/2))*a*c-393*I*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-315*ln((c*a*tan(f*x+e)+(c*a*(1
+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c-259*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*ta
n(f*x+e)^2+164*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(tan(f*x+e)+I)^3/(c*a)^(1/2)/(c*a*(1+tan(f*x+e)^2))^(
1/2)

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maxima [B]  time = 0.81, size = 919, normalized size = 3.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-((1260*a^4*cos(4*f*x + 4*e) + 2520*a^4*cos(2*f*x + 2*e) + 1260*I*a^4*sin(4*f*x + 4*e) + 2520*I*a^4*sin(2*f*x
+ 2*e) + 1260*a^4)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) + 1) + (1260*a^4*cos(4*f*x + 4*e) + 2520*a^4*cos(2*f*x + 2*e) + 1260*I*a^4*sin(4*f*x +
4*e) + 2520*I*a^4*sin(2*f*x + 2*e) + 1260*a^4)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))),
-sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (192*a^4*cos(4*f*x + 4*e) + 384*a^4*cos(2*f*x + 2
*e) + 192*I*a^4*sin(4*f*x + 4*e) + 384*I*a^4*sin(2*f*x + 2*e) - 744*a^4)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) - (1728*a^4*cos(4*f*x + 4*e) + 3456*a^4*cos(2*f*x + 2*e) + 1728*I*a^4*sin(4*f*x + 4*e) + 3456*
I*a^4*sin(2*f*x + 2*e) + 2520*a^4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-630*I*a^4*cos(4*f*
x + 4*e) - 1260*I*a^4*cos(2*f*x + 2*e) + 630*a^4*sin(4*f*x + 4*e) + 1260*a^4*sin(2*f*x + 2*e) - 630*I*a^4)*log
(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (630*I*a^4*cos(4*f*x + 4*e) + 1260*I*a^4*co
s(2*f*x + 2*e) - 630*a^4*sin(4*f*x + 4*e) - 1260*a^4*sin(2*f*x + 2*e) + 630*I*a^4)*log(cos(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (-192*I*a^4*cos(4*f*x + 4*e) - 384*I*a^4*cos(2*f*x + 2*e) + 192*a^4
*sin(4*f*x + 4*e) + 384*a^4*sin(2*f*x + 2*e) + 744*I*a^4)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
- (1728*I*a^4*cos(4*f*x + 4*e) + 3456*I*a^4*cos(2*f*x + 2*e) - 1728*a^4*sin(4*f*x + 4*e) - 3456*a^4*sin(2*f*x
+ 2*e) + 2520*I*a^4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-72*I*c^2*cos(4*
f*x + 4*e) - 144*I*c^2*cos(2*f*x + 2*e) + 72*c^2*sin(4*f*x + 4*e) + 144*c^2*sin(2*f*x + 2*e) - 72*I*c^2)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^(9/2)/(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(9/2)/(c - c*tan(e + f*x)*1i)^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(9/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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